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I know all the rules but I don't know how make assumptions and build a proof in general.

Example : prove : $((A \rightarrow B) \wedge(C \rightarrow B))\rightarrow(A \wedge C)\rightarrow B$

I'm only able to do these 2 steps :

$ \frac{(A \wedge B)}{B} \frac{(A ),( C)}{(A \wedge C)}$

I don't know how to build the $\rightarrow$'s using the $\rightarrow$ introduction rule.

Let's try it again :

$1. [(A \rightarrow B)\wedge(C \rightarrow B)] $

$2. (A \wedge C),B $

$3. B $ is it correct?

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  • $\begingroup$ Maybe you must add a couple of parenthesis to improve readibility. $\endgroup$ – Mauro ALLEGRANZA Feb 6 '18 at 14:29
  • $\begingroup$ Start assuming the antecedent: $((A \to B) \land (C \to B))$. $\endgroup$ – Mauro ALLEGRANZA Feb 6 '18 at 14:30
  • $\begingroup$ Then assume $(A \land C)$. $\endgroup$ – Mauro ALLEGRANZA Feb 6 '18 at 14:30
  • $\begingroup$ $1). (A \rightarrow B)\wedge(C \rightarrow B) 2). (A \wedge C),B 3). B $ is it correct? $\endgroup$ – Bleeeaa Feb 6 '18 at 14:34
  • $\begingroup$ No, it is not.... $\endgroup$ – Mauro ALLEGRANZA Feb 6 '18 at 14:39
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We have to prove an implication, so we start by assuming the antecedent:

  1. $((A \to B) \land (C \to B))$ (ass.)

We have to prove an implication from this so again we assume the antecedent:

  1. $(A \land C)$ (ass)
  2. $A \to B$ from 1. and elimination of $\land$.
  3. $A$ from 2. and elimination of $\land$ again.
  4. $B$ by modus ponens from 3 and 4.
  5. $(A \land C) \to B$ from introduction $\to$ (cancel ass. 2)

Now your required statement follows from ass 1 (which will be eliminated) and again introduction of $\to$.

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